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Year 10 Maths - 10A Linear Relations

Further Applications of Simultaneous Equations

20 practice questions 1 video lesson Theory + worked examples
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Theory

A word problem with two unknowns needs two facts. Name each unknown with a pronumeral, turn each fact into an equation, solve the pair by substitution or elimination, and answer in words. This Year 10A maths page shows how to translate ages, digits, fractions, mixtures, percentages and interest, fixed charges, speeds, travel times, geometry facts and curves through given points into equations, and how to check the answer against the story.

Modelling a problem means writing equations that say the same thing as the words. Each different fact in the question gives one equation. Two unknowns need two independent equations.

Define each pronumeral fully, with its unit and time: "let \(x\) be the price of an adult ticket, in dollars, on Friday". Vague definitions cause most errors.

After solving, interpret: does the answer make sense (a whole number of people, a positive speed, a digit from \(0\) to \(9\))? Does it fit every fact in the question?

If the two facts turn out to be the same equation in disguise (one is a multiple of the other), there are infinitely many solutions, and the problem does not fix a single answer.

Closing speedsTwo travellers with speeds u and v. Moving towards each other, the gap between them closes at u + v. Moving the same way with the faster one behind, the gap closes at u - v. Towards each other u v gap closes at u + v Same direction (faster one behind) u v gap closes at u − v
For travel problems, add the speeds when the travellers approach each other and subtract when one chases the other. Then distance \(=\) speed \(\times\) time.
Fixed fee plus a rateA plumber's charge C = F + rh drawn as a line. Two jobs, 2 hours for 230 dollars and 5 hours for 470 dollars, are points on the line. The y-intercept is the fixed fee of 70 dollars and the gradient is the hourly rate of 80 dollars. hours dollars 1 2 3 4 5 6 100 200 300 400 500 (2, 230) (5, 470) fixed fee slope = rate 0
Two jobs give two equations: \(F + 2r = 230\) and \(F + 5r = 470\). Subtracting gives \(3r = 240\), so \(r = 80\) and \(F = 70\).

Common translations:

WordsAlgebra
two-digit number, tens digit \(t\), units digit \(u\)\(10t + u\); reversed \(10u + t\)
a fraction \(\dfrac{a}{b}\) with \(2\) added to top and bottom\(\dfrac{a + 2}{b + 2}\)
\(15\%\) more than \(x\)\(1.15x\)
\(x\) dollars invested at \(5\%\) p.a. for one yearinterest \(0.05x\)
fixed charge \(F\) plus \(r\) per unit, \(n\) units\(F + rn\)
time to travel \(d\) km at \(v\) km/h\(\dfrac{d}{v}\)
two machines at \(p\) and \(q\) units per hour for \(h\) hours\(h(p + q)\)
\[\text{distance} = \text{speed} \times \text{time}\]\[\text{amount} = \text{rate} \times \text{time}\]
d=vt
Shapes. Geometry facts give equations too: opposite angles of a parallelogram are equal, co-interior angles add to \(180^\circ\), angles in a triangle add to \(180^\circ\). A curve \(y = ax^2 + bx\) through two given points gives two equations in \(a\) and \(b\).
Messy coefficients. Multiply decimals out first (\(\times 100\) for percentages), divide by any common factor, and use elimination: \(250A + 150C = 9\,500\) simplifies to \(5A + 3C = 190\).

Solving a word problem

  1. Read to the end. Find the two unknowns the question asks about (or that lead to the answer).
  2. Define a pronumeral for each, with units: "let \(b\) km/h be the boat's speed in still water".
  3. Write one equation for each fact. A table (people, now, in \(5\) years; or amount, price, value) helps organise the facts.
  4. Simplify each equation, then solve by substitution (if one is \(x = \ldots\)) or elimination.
  5. Check in the words of the question, not just in your equations.
  6. Answer in a sentence with units, and answer the part asked (this year's number, the total bill, the time together).

Tips for common types

  1. Ages: add or subtract the same years for everyone.
  2. Mixtures and tickets: one equation for quantities, one for money.
  3. Before and after a change: let the pronumerals be the original values, then multiply by the percentage factor.
  4. Working together: find each rate (litres per hour), then time \(=\) amount \(\div\) combined rate.
  5. Not every question is a story: equations with \(\dfrac{1}{x}\) and \(\dfrac{1}{y}\) (let \(u = \dfrac{1}{x}\), \(v = \dfrac{1}{y}\)), letters as coefficients (eliminate, then factorise and divide) and unknown coefficients from a known point (substitute the point to get two equations) use the same methods as the Substitution and Elimination pages.
  6. No solution or many: if eliminating gives \(0 = 5\), the facts contradict each other; \(0 = 0\) means one fact repeats the other.
Example 1 — Ages
In \(4\) years, Mia's mother will be three times as old as Mia. Six years ago she was eight times as old as Mia. How old are they now?
Solution

Let Mia be \(a\) years old and her mother \(m\) years old now. Add or subtract the same number of years for both people.

\(m + 4\)\(=\)\(3(a + 4)\)
\(m\)\(=\)\(3a + 8 \quad (1)\)
\(m - 6\)\(=\)\(8(a - 6)\)
\(m\)\(=\)\(8a - 42 \quad (2)\)

Both equations give \(m\), so set them equal.

\(8a - 42\)\(=\)\(3a + 8\)
\(5a\)\(=\)\(50\)
\(a\)\(=\)\(10\)
\(m\)\(=\)\(3(10) + 8 = 38\)

Check: in \(4\) years they are \(14\) and \(42 = 3 \times 14\ \checkmark\); six years ago \(4\) and \(32 = 8 \times 4\ \checkmark\)

Mia is \(\mathbf{10}\) and her mother is \(\mathbf{38}\).

m+4=3(a+4)
Example 2 — A mixture
A café blends coffee beans costing \(\$24\) per kg with beans costing \(\$36\) per kg to make \(30\) kg of a blend worth \(\$28\) per kg. How much of each does it use?
Solution

Let \(x\) kg be the \(\$24\) beans and \(y\) kg the \(\$36\) beans. One equation counts kilograms, the other counts dollars.

\(x + y\)\(=\)\(30 \quad (1)\)
\(24x + 36y\)\(=\)\(28 \times 30\)
\(\)\(=\)\(840 \quad (2)\)

\((1) \times 24\):

\(24x + 24y\)\(=\)\(720 \quad (3)\)

\((2) - (3)\):

\(12y\)\(=\)\(120\)
\(y\)\(=\)\(10\)
\(x\)\(=\)\(20\)

Check: \(24(20) + 36(10)\) \(= 480 + 360 = 840\ \checkmark\)

It uses \(\mathbf{20}\) kg of the \(\$24\) beans and \(\mathbf{10}\) kg of the \(\$36\) beans.

24x+36y=840
Example 3 — Speed with a current
A boat travels \(36\) km downstream in \(2\) hours and makes the same trip back upstream in \(3\) hours. Find the speed of the boat in still water and the speed of the current.
Solution

Let the boat's still-water speed be \(b\) km/h and the current \(c\) km/h. Downstream the current helps (\(b + c\)); upstream it slows the boat (\(b - c\)). Use distance \(=\) speed \(\times\) time.

\(2(b + c)\)\(=\)\(36\)
\(b + c\)\(=\)\(18 \quad (1)\)
\(3(b - c)\)\(=\)\(36\)
\(b - c\)\(=\)\(12 \quad (2)\)

\((1) + (2)\):

\(2b\)\(=\)\(30\)
\(b\)\(=\)\(15\)
\(c\)\(=\)\(18 - 15 = 3\)

Check upstream: \(3 \times (15 - 3) = 36\ \checkmark\)

Boat \(\mathbf{15}\) km/h, current \(\mathbf{3}\) km/h.

b+c=18,b−c=12
Example 4 — Percentage changes
Last month a café sold \(500\) coffees and teas altogether. This month coffee sales rose by \(10\%\) and tea sales fell by \(20\%\), and it sold \(490\) drinks.
(i) How many of each were sold last month?
(ii) How many coffees were sold this month?
Solution

(i) Let \(c\) coffees and \(t\) teas be last month's sales. A \(10\%\) rise multiplies by \(1.1\); a \(20\%\) fall multiplies by \(0.8\).

\(c + t\)\(=\)\(500 \quad (1)\)
\(1.1c + 0.8t\)\(=\)\(490 \quad (2)\)

\((1) \times 0.8\):

\(0.8c + 0.8t\)\(=\)\(400 \quad (3)\)

\((2) - (3)\):

\(0.3c\)\(=\)\(90\)
\(c\)\(=\)\(300\)
\(t\)\(=\)\(200\)

Last month: \(\mathbf{300}\) coffees and \(\mathbf{200}\) teas. Check: \(330 + 160 = 490\ \checkmark\)

(ii) This month: \(1.1 \times 300 = \mathbf{330}\) coffees.

1.1c+0.8t=490

Answer the question asked: part (ii) wants this month's number, not \(c\).

Common pitfalls

Define the pronumerals precisely. "Let \(x\) = boys" is unclear. Write "let \(x\) be the number of boys last year".
Change both ages. "In \(5\) years" means \(m + 5\) and \(a + 5\), not just one of them.
A percentage of the original. A \(20\%\) fall from \(t\) is \(0.8t\). Do not subtract \(20\) or \(0.2\).
Keep units consistent. Mix minutes and hours, or cents and dollars, and the equations are wrong. Convert first.
Answer the actual question. Finding \(x\) and \(y\) is not the end if the question asks for a total, a difference or a later value.

Frequently asked questions

How do you set up simultaneous equations from a word problem?

Choose a pronumeral for each unknown and say exactly what it stands for. Then turn each separate fact in the question into an equation. Two unknowns need two facts.

Should I use substitution or elimination for word problems?

Use substitution when one equation is already in the form x equals something, such as in age problems. Use elimination when both equations look like ax plus by equals c, such as in mixture or ticket problems.

How do you solve age problems with simultaneous equations?

Let the pronumerals be the ages now. For a time in the future add the same number of years to both ages; for the past subtract them. Write one equation for each statement and solve.

How do you handle percentage increases in these problems?

Let the pronumerals be the original amounts. An increase of 10 percent multiplies by 1.1 and a decrease of 20 percent multiplies by 0.8. Multiply through by 10 or 100 to clear the decimals.

How do I check my answer?

Put your values back into the words of the problem and check every fact, not just the equations, in case an equation was set up incorrectly.